Problem: Let $y=\tan(x^2-4x)$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{\cos^2(x^2-4x)}$ (Choice B) B $\dfrac{2x-4}{\cos^2(x^2-4x)}$ (Choice C) C $\dfrac{1}{\sin^2(x^2-4x)}$ (Choice D) D $\dfrac{2x-4}{\sin^2(x^2-4x)}$
$\tan(x^2-4x)$ is a trigonometric expression, but its argument isn't simply $x$. Therefore, it defines a composite trigonometric function. In other words, suppose $u(x)=x^2-4x$, then $y=\tan\Bigl(u(x)\Bigr)$. $\dfrac{dy}{dx}$ can be found using the following identity: $\dfrac{d}{dx}\left[\tan\Bigl(u(x)\Bigr)\right]=\dfrac{u'(x)}{\cos^2\Bigl(u(x)\Bigr)}$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\tan(x^2-4x) \\\\ &=\dfrac{d}{dx}\tan\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=x^2-4x} \\\\ &=\dfrac{u'(x)}{\cos^2\Bigl(u(x)\Bigr)} \\\\ &=\dfrac{2x-4}{\cos^2\Bigl(x^2-4x\Bigr)}&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{2x-4}{\cos^2\Bigl(x^2-4x\Bigr)}$.